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zalisa [80]
3 years ago
9

The Greek alphabet has 24 distinct lowercase letters. How many bits are needed to be able to encode any single lowercase Greek l

etter
Engineering
1 answer:
Alexus [3.1K]3 years ago
4 0

Answer:

You need 5 bits to encode al the 24 greek distinct lowercase letters.

Explanation:

o encode 24 greek distinct lowercase letters, you should be able to write in binary language 24 different combinations.

With 1 bit you have only 2 possible combinations. Not enough.

With 2 bit you have only 4 possible combinations. Not enough.

With 3 and 4 bit you have only 8 and 16 possible combinations. Not enough.

With 5 bit you have 32 possible combinations. This is enough to encode all the 24 Greek letters.

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A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe
ankoles [38]

Answer:

F =  8849 N

Explanation:

Given:

Load at a given point = F =  4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ_{fs} = F_{f} L / \pi  R^{3}

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π  ( 5.6 x 10⁻³ ) ³

  = 4250 ( 44 x 10⁻³ )  / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 4250 ( 11 / 250  ) / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

  = 187 / 3.141593 (0.0056)³

  = 338943767.745358

  = 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

F_{f} = 2 σ_{fs} d³/3 L

F = 2σd³/3L

  = 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

  = 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 676000000 ( 12  x  1/1000  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  3  / 250  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  27  / 15625000 )  / 3 ( 44 x 10⁻³)

  = 146016  / 125 / 3 ( 44 x 1 / 1000  )

  = ( 146016  / 125 ) /  (3 ( 11 /  250 ))

  =  97344  / 11

F =  8849 N

4 0
3 years ago
Am I alive I really need to know?
Nesterboy [21]
Hell no,cause i’m not
5 0
3 years ago
The position of a particle is given by s = 0.27t
Natali [406]
Sorry bro people do this22.2 pls
8 0
2 years ago
A cylindrical metal specimen of initial diameter d0 =14 mm, initial length L0=53 mm, strain hardening exponent n=0.31, strength
Marrrta [24]

Answer:

a) Ef = 0.755

b) length of specimen( Lf )= 72.26mm

  diameter at fracture = 9.598 mm

c) max load ( Fmax ) = 52223.24 N

d) Ft = 51874.67 N

Explanation:

a) Determine the true strain at maximum load and true strain at fracture

True strain at maximum load

Df = 9.598 mm

True strain at fracture

Ef = 0.755

b) determine the length of specimen at maximum load and diameter at fracture

Length of specimen at max load

Lf = 72.26 mm

Diameter at fracture

= 9.598 mm

c) Determine max load force

Fmax = 52223.24 N

d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test

F = 51874.67 N

attached below is a detailed solution of the question above

3 0
3 years ago
What do you own that might not be manufactured?
horrorfan [7]

Answer:

A pet

Explanation:

Latin time I checked animals aren't made by people? I honestly don't know if this helps but I'm technically not wrong.

8 0
3 years ago
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